Periodic table
Periodic table
You are encouraged to solve this task according to the task description, using any language you may know.
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Display the row and column in the periodic table of the given atomic number.
- The periodic table
Let us consider the following periodic table representation.
__________________________________________________________________________ | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 | | | |1 H He | | | |2 Li Be B C N O F Ne | | | |3 Na Mg Al Si P S Cl Ar | | | |4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr | | | |5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe | | | |6 Cs Ba * Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn | | | |7 Fr Ra ° Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og | |__________________________________________________________________________| | | | | |8 Lantanoidi* La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu | | | |9 Aktinoidi° Ak Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr | |__________________________________________________________________________|
- Example test cases;
-
1
->1 1
-
2
->1 18
-
29
->4 11
-
42
->5 6
-
57
->8 4
-
58
->8 5
-
72
->6 4
-
89
->9 4
Note that atomic values corresponding to Lantanoidi (57 to 71) and Aktinoidi (89 to 103) are to be mapped in their own rows.
- See also
- the periodic table
- This task was an idea from CompSciFact
- The periodic table in ascii that was used as template
Python
A solution trying hard not to encode too much data about the table.
<lang Python> def perta(atomic) -> (int, int):
NOBLES = 2, 10, 18, 36, 54, 86, 118 INTERTWINED = 0, 0, 0, 0, 0, 57, 89 INTERTWINING_SIZE = 14 LINE_WIDTH = 18
prev_noble = 0 for row, noble in enumerate(NOBLES): if atomic <= noble: # we are at the good row. We now need to determine the column nb_elem = noble - prev_noble # number of elements on that row rank = atomic - prev_noble # rank of the input element among elements if INTERTWINED[row] and INTERTWINED[row] <= atomic <= INTERTWINED[row] + INTERTWINING_SIZE: # lantanides or actinides row += 2 col = rank + 1 else: # not a lantanide nor actinide # handle empty spaces between 1-2, 4-5 and 12-13. nb_empty = LINE_WIDTH - nb_elem # spaces count as columns inside_left_element_rank = 2 if noble > 2 else 1 col = rank + (nb_empty if rank > inside_left_element_rank else 0) break prev_noble = noble return row+1, col
- small test suite
TESTS = {
1: (1, 1), 2: (1, 18), 29: (4,11), 42: (5, 6), 58: (8, 5), 59: (8, 6), 57: (8, 4), 71: (8, 18), 72: (6, 4), 89: (9, 4), 90: (9, 5), 103: (9, 18),
}
for input, out in TESTS.items():
found = perta(input) print('TEST:{:3d} -> '.format(input) + str(found) + (f' ; ERROR: expected {out}' if found != out else ))
</lang>